R help - Jan 2009 - lm: how are polynomial functions interpreted?

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On Mon, 12 Jan 2009, cgw at witthoft.com wrote:

[nothing deleted]

matplot(1:100, lm(rnorm(100)~poly(1:100,4),x=T)$x ) # for example
>
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Ahem!
> and provide commented, minimal, self-contained, reproducible code.
......^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

damn.
My apologies to everyone -- I sent one message and it got destroyed somehow.

Here's the part that was missing, which led to rather a lot of confusion 
on all parts.

The two recent responses to a question about lm suggested

1)  lm(y~poly(x,2))

2) lm(y~I(x^2))

So my question was *supposed* to be related to how lm operated 
differently (if at all) on these two different 'versions' of a quadratic
fit.  My gut reaction was that y~I(x^2) would not be the same as y~f(x) 
where f(x) is a+bx+cx^2 .

So what I was trying to find out was just how lm() deals with various 
definitions of the orthogonal polynomials its presented with.  Another 
way, maybe, to ask, is: how does one specify to fit exactly to

   a + bx +cx^2  vs

   bx + cx^2   vs
   cx^2
?

Thanks and apologies again to all the people who quite properly 
misunderstood what I was harping on due to the munging of my first post.

Carl


David Winsemius wrote:
> 
> On Jan 12, 2009, at 5:57 PM, Carl Witthoft wrote:
> 
>> Well..... *_* ,
>>
>> I think it should have been clear that this was not a question for 
>> which any code exists.  In fact, I gave two very specific examples of 
>> function calls.
> 
> Huh? I think most of the readers of the list saw a basically empty 
> message body. That was the point of Berry's "[nothing
deleted]". If you
> are under the belief that this is a continuation of an earlier question, 
> then it may not be threading up in the manner you hoped for.
>

A 
http://polynomial-trimonial-binomial.blogspot.com/2011/10/how-to-deal-with-polynomials.html
polynomial function  is evaluated by foiling out the equation so you can
solve for 'x'.

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